Let's break down and solve each of the three problems from the image.

Problem 1:

Given two sides $a = 3.2$ and $b = 4.6$ of a triangle, and the angle $\theta = 60^\circ$ between the two sides, find the length of the third side and the area of the triangle.

1. Finding the third side (using the cosine rule):

   The cosine rule is: $c^2 = a^2 + b^2 - 2ab\cos(\theta)$ Substituting the given values: $c^2 = 3.2^2 + 4.6^2 - 2 \times 3.2 \times 4.6 \times \cos(60^\circ)$ Since $\cos(60^\circ) = 0.5$: $c^2 = 10.24 + 21.16 - 2 \times 3.2 \times 4.6 \times 0.5$ $c^2 = 31.4 - 14.72 = 16.68$ $c = \sqrt{16.68} \approx 4.08$ So, the third side $c \approx 4.08 \, \text{cm}$.

2. Finding the area of the triangle (using the formula for area with two sides and the included angle):

   The area $A$ of a triangle is: $A = \frac{1}{2}ab\sin(\theta)$ Substituting the given values: $A = \frac{1}{2} \times 3.2 \times 4.6 \times \sin(60^\circ)$ Since $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$: $A = 0.5 \times 3.2 \times 4.6 \times 0.866 \approx 6.38 \, \text{cm}^2$ So, the area of the triangle is approximately $6.38 \, \text{cm}^2$.

Problem 2:

Given the radius $r = 5 \, \text{cm}$ of a circle, find its circumference and area.

1. Circumference of the circle:

   The formula for the circumference $C$ of a circle is: $C = 2\pi r$ Substituting $r = 5 \, \text{cm}$: $C = 2 \times \pi \times 5 \approx 31.42 \, \text{cm}$

2. Area of the circle:

   The formula for the area $A$ of a circle is: $A = \pi r^2$ Substituting $r = 5 \, \text{cm}$: $A = \pi \times 5^2 = 25\pi \approx 78.54 \, \text{cm}^2$ So, the area of the circle is approximately $78.54 \, \text{cm}^2$.

Problem 3:

Given the base of a right-angled triangle $= 5 \, \text{cm}$ and the acute angle $\theta = 42^\circ$, find the length of the perpendicular.

In a right-angled triangle, the relationship between the sides and angles can be found using trigonometric ratios. Here, we will use the tangent function: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{perpendicular}}{\text{base}}$ Substituting the known values: $\tan(42^\circ) = \frac{\text{perpendicular}}{5}$ Using $\tan(42^\circ) \approx 0.9004$: $0.9004 = \frac{\text{perpendicular}}{5}$ $\text{perpendicular} = 0.9004 \times 5 \approx 4.50 \, \text{cm}$ So, the length of the perpendicular is approximately $4.50 \, \text{cm}$.

Would you like any more details or further explanations? Here are 5 additional questions related to the topics above:

1. How do we generalize the cosine rule for non-obtuse triangles?
2. Can the formula for the area of a triangle using sine be derived from basic geometry principles?
3. How does the formula for the circumference of a circle change with different radii?
4. Can you use the sine or cosine functions to find the height of the triangle in Problem 3?
5. How does the unit circle relate to trigonometric values like $\sin(60^\circ)$ and $\tan(42^\circ)$?

Tip: When solving problems involving angles in triangles, remember to always check if the triangle is right-angled for simpler trigonometric solutions.